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# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.2

• Last Updated : 28 Dec, 2020

### Question 1: Write the following in the expanded form:

(i) (a + 2b + c)2

(ii) (2a âˆ’ 3b âˆ’ c)2

(iii) (âˆ’3x+y+z)2

(iv) (m+2nâˆ’5p)2

(v) (2+xâˆ’2y)2

(vi) (a2 +b2 +c2)2

(vii) (ab+bc+ca)2

(viii) (x/y+y/z+z/x)2

(ix) (a/bc + b/ac + c/ab)2

(x) (x+2y+4z)2

(xi) (2xâˆ’y+z)2

(xii) (âˆ’2x+3y+2z)2

Solution:

By following the identity,

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz

(i) (a + 2b + c)2

= a2 + (2b)2 + c2 + 2a(2b) + 2ac + 2(2b)c

= a2 + 4b2 + c2 + 4ab + 2ac + 4bc

(ii) (2a âˆ’ 3b âˆ’ c)2

= [(2a) + (âˆ’3b) + (âˆ’c)]2

= (2a)2 + (âˆ’3b)2 + (âˆ’c)2 + 2(2a)(âˆ’3b) + 2(âˆ’3b)(âˆ’c) + 2(2a)(âˆ’c)

= 4a2 + 9b2 + c2 âˆ’ 12ab + 6bc âˆ’ 4ca

(iii) (âˆ’3x+y+z)2

= [(âˆ’3x)2 + y2 + z2 + 2(âˆ’3x)y + 2yz + 2(âˆ’3x)z]

= 9x2 + y2 + z2 âˆ’ 6xy + 2yz âˆ’ 6xz

(iv) (m+2nâˆ’5p)2

= m2 + (2n)2 + (âˆ’5p)2 + 2m Ã— 2n + (2Ã—2nÃ—âˆ’5p) + 2m Ã— (âˆ’5p)

= m2 + 4n2 + 25p2 + 4mn âˆ’ 20np âˆ’ 10pm

(v) (2+xâˆ’2y)2

= 22 + x2 + (âˆ’2y) 2 + 2(2)(x) + 2(x)(âˆ’2y) + 2(2)(âˆ’2y)

= 4 + x2 + 4y2 + 4 x âˆ’ 4xy âˆ’ 8y

(vi) (a2 +b2 +c2)2

= (a2)2 + (b2)2 + (c2)2 + 2a2 b2 + 2b2c2 + 2a2c2

= a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2

(vii) (ab+bc+ca)2

= (ab)2 + (bc)2 + (ca)2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)

= a2b2 + b2c2 + c2a2 + 2(ac)b2 + 2(ab)(c)2 + 2(bc)(a)2

(viii) (x/y+y/z+z/x)2

(ix) (a/bc + b/ac + c/ab)2

(x) (x+2y+4z)2

= x2 + (2y)2 + (4z)2 + (2x)(2y) + 2(2y)(4z) + 2x(4z)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(xi) (2xâˆ’y+z)2

= (2x)2 + (âˆ’y)2 + (z)2 + 2(2x)(âˆ’y) + 2(âˆ’y)(z) + 2(2x)(z)

= 4x2 + y2 + z2 âˆ’ 4xy âˆ’ 2yz + 4xz

(xii) (âˆ’2x+3y+2z)2

= (âˆ’2x)2 + (3y)2 + ( 2z)2 + 2(âˆ’2x)(3y) + 2(3y)(2z) + 2(âˆ’2x)(2z)

= 4x2 + 9y2 + 4z2 âˆ’12xy + 12yz âˆ’8xz

### Question 2: Simplify

(i) (a + b + c)2 + (a âˆ’ b + c)2

(ii) (a + b + c)2 âˆ’ (a âˆ’ b + c)2

(iii) (a + b + c)2 + (a â€“ b + c)2 + (a + b âˆ’ c)2

(iv) (2x + p âˆ’ c)2 âˆ’ (2x âˆ’ p + c)2

(v) (x2 + y2 âˆ’ z2)2 âˆ’ (x2 âˆ’ y2 + z2)2

Solution:

(i) (a + b + c)2 + (a âˆ’ b + c)2

= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (âˆ’b)2 + c2 âˆ’2abâˆ’2bc+2ca)

= 2a2 + 2 b2 + 2c2 + 4ca

(ii) (a + b + c)2 âˆ’ (a âˆ’ b + c)2

= (a2 + b2 + c2 + 2ab+2bc+2ca) âˆ’ (a2 + (âˆ’b)2 + c2 âˆ’2abâˆ’2bc+2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca âˆ’ a2 âˆ’ b2 âˆ’ c2 + 2ab + 2bc âˆ’ 2ca

= 4ab + 4bc

(iii) (a + b + c)2 + (a â€“ b + c)2 + (a + b âˆ’ c)

= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c)2 âˆ’ 2ab âˆ’ 2cb + 2ca) + (a2 + b2 + c2 + 2ab âˆ’ 2bc â€“ 2ca)

= 3a2 + 3b2 + 3c2 + 2ab âˆ’ 2bc + 2ca

(iv) (2x + p âˆ’ c)2 âˆ’ (2x âˆ’ p + c)2

= [4x2 + p2 + c2 + 4xp âˆ’ 2pc âˆ’ 4xc] âˆ’ [4x2 + p2 + c2 âˆ’ 4xpâˆ’ 2pc + 4xc]

= 4x2 + p2 + c2 + 4xp âˆ’ 2pc âˆ’ 4cx âˆ’ 4x2 âˆ’ p2 âˆ’ c2 + 4xp + 2pcâˆ’ 4cx

= 8xp âˆ’ 8xc

= 8(xp âˆ’ xc)

(v) (x2 + y2 âˆ’ z2)2 âˆ’ (x2 âˆ’ y2 + z2)2

= (x2 + y2 + (âˆ’z)2)2 âˆ’ (x2 âˆ’ y2 + z2)2

= [x4 + y4 + z4 + 2x2y2 â€“ 2y2z2 â€“ 2x2z2 âˆ’ [x4 + y4 + z4 âˆ’ 2x2y2 âˆ’ 2y2z2 + 2x2z2]

= 4x2y2 â€“ 4z2x2

### Question 3: If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.

Solution:

Given,

a + b + c = 0 and a2 + b2 + c2 = 16

Choose a + b + c = 0

Squaring both sides,

(a + b + c)2 = 0

a2 + b2 + c2 + 2(ab + bc + ca) = 0

16 + 2(ab + bc + c) = 0

2(ab + bc + ca) = -16

ab + bc + ca = -16/2 = -8

or

ab + bc + ca = -8

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